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A sample of water has a hardness expressed as 80 ppm of ${Ca}^{2 +}$ . This sample is passed through an ion exchange column and the ${Ca}^{2 +}$ is replaced by $H^{+}$ . What is the pH of the water after it has been so treated? [Atomic mass of Ca = 40]

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2.4

5.4

2.7

answer is D.

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Detailed Solution

According to charge balance ${eq}^{n}, [{Ca}^{2 +}] = 2 [H^{+}]$

$80 g {Ca}^{2 +}$ in $10^{6} g H_{2} O$

Each ${Ca}^{2 +}$ will be replaced by $2 H^{+}$

So, 4 mole $H^{+}$ in $10^{6} ml H_{2} O$

$\begin{array}{l} [H^{+}] = 4 \times 10^{- 3} \\ pH = 3 \log 4 = 2.4 \end{array}$

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